hdoj 3820 Golden Eggs 【双二分图构造最小割模型】

2
2 2 100 100
1 1
5 1
1 4
1 1
1 4 85 95
100 100 10 10
10 10 100 100

Case 1: 9
Case 2: 225

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 6000
#define MAXM 50000
#define INF 0x3f3f3f3f
using namespace std;
struct Edge
{
    int from, to, cap, flow, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int dist[MAXN], cur[MAXN];
bool vis[MAXN];
int N, M, G, S;
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
    Edge E1 = {u, v, w, 0, head[u]};
    edge[edgenum] = E1;
    head[u] = edgenum++;
    Edge E2 = {v, u, 0, 0, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum++;
}
int point(int x, int y)
{
    return (x-1) * M + y;
}
bool judge(int x, int y)
{
    return x >= 1 && x <= N && y >= 1 && y <= M;
}
int sum;//金蛋 银蛋的总价值
int source, sink;
void getMap()
{
    source = 0, sink = N * M * 2 + 1;
    sum = 0;
    int a;
    for(int i = 1; i <= N; i++)
    {
        for(int j = 1; j <= M; j++)
        {
            scanf("%d", &a);
            sum += a;
            if((i+j) & 1)//源点 连 S1集
                addEdge(source, point(i, j), a);
            else//T1集 连汇点
                addEdge(point(i, j), sink, a);
        }
    }
    for(int i = 1; i <= N; i++)
    {
        for(int j = 1; j <= M; j++)
        {
            scanf("%d", &a);
            sum += a;
            if((i+j) & 1)//S2集 连汇点
                addEdge(point(i, j) + N*M, sink, a);//S2集合 x 格子的编号为S1集合 x 格子编号 + N*M
            else//源点连 T2集
                addEdge(source, point(i, j) + N*M, a);
        }
    }
    int move[4][2] = {0,1, 0,-1, 1,0, -1,0};
    for(int i = 1; i <= N; i++)
    {
        for(int j = 1; j <= M; j++)
        {
            for(int k = 0; k < 4; k++)
            {
                int x = i + move[k][0];
                int y = j + move[k][1];
                if(!judge(x, y)) continue;
                if((i+j) & 1)//S1集
                {
                    addEdge(point(i, j), point(i, j) + N*M, INF);//S1集 向 S2集建边无穷大 表示不能同时选
                    if((x+y) % 2 == 0)
                        addEdge(point(i, j), point(x, y), G);//S1集 向 T1集建边
                }
                else//T1集
                {
                    addEdge(point(i, j) + N*M, point(i, j), INF);//T2集 向 T1集建边无穷大 表示不能同时选
                    if((x+y) & 1)
                        addEdge(point(i, j)+N*M, point(x, y) + N*M, S);//T2集 向 S2集建边 
                }
            }
        }
    }
}
bool BFS(int s, int t)
{
    queue<int> Q;
    memset(dist, -1, sizeof(dist));
    memset(vis, false, sizeof(vis));
    dist[s] = 0;
    vis[s] = true;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            Edge E = edge[i];
            if(!vis[E.to] && E.cap > E.flow)
            {
                dist[E.to] = dist[u] + 1;
                if(E.to == t) return true;
                vis[E.to] = true;
                Q.push(E.to);
            }
        }
    }
    return false;
}
int DFS(int x, int a, int t)
{
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1; i = edge[i].next)
    {
        Edge &E = edge[i];
        if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap-E.flow), t)) > 0)
        {
            edge[i].flow += f;
            edge[i^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}
int Maxflow(int s, int t)
{
    int flow = 0;
    while(BFS(s, t))
    {
        memcpy(cur, head, sizeof(head));
        flow += DFS(s, INF, t);
    }
    return flow;
}
int main()
{
    int t, k = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d%d%d", &N, &M, &G, &S);
        init();
        getMap();
        printf("Case %d: %d\n", k++, sum - Maxflow(source, sink));
    }
    return 0;
}